document.write( "Question 94296: square root(2x-1) - square root(x+3) = 1
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Algebra.Com's Answer #68645 by stanbon(75887)\"\" \"About 
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square root(2x-1) - square root(x+3) = 1
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\n" ); document.write( "Square both sides to get:
\n" ); document.write( "(2x-1) -2sqrt[(2x-1)(x+3)]+(x+3) = 1
\n" ); document.write( "Simplify:
\n" ); document.write( "-2sqrt[2x^2+5x-3)] = -3x-1
\n" ); document.write( "Square both sides to get:
\n" ); document.write( "4(2x^2+5x-3) = 9x^2+6x+1
\n" ); document.write( "Simplify:
\n" ); document.write( "x^2-14x+13 = 0
\n" ); document.write( "(x-1)(x-13)=0
\n" ); document.write( "x = 1 or x=13
\n" ); document.write( "Checking for extraneous roots in \"square root(2x-1) - square root(x+3) = 1\":
\n" ); document.write( "If x = 1: sqrt(1) - sqrt(4) = 1
\n" ); document.write( "1-2 = 1
\n" ); document.write( "Wrong, so x=1 is not a solution of the orignial equation.
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\n" ); document.write( "If x= 13 : sqrt(25) - sqrt(16) = 1
\n" ); document.write( "5-4 = 1
\n" ); document.write( "True, so x=13 is the final solution.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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